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In QM, we use tensor products to construct the vector space of the states of a multi-particle system - but that construction doesn’t seem to have a counterpart in classical mechanics. In QM, it seems to be required to be able to represent entangled states.

Is it considered a “postulate” about how to represent the states of multi-particle systems in QM? Is it correct that it does not have an equivalent in classical mechanics, where we are content to use direct sums of vector spaces?

Qmechanic
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Frank
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3 Answers3

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Yes, this is one of the postulates of quantum mechanics. For example, see section 2.2 of Nielsen and Chuang, Quantum Information and Quantum Computation, where this is postulate 4.

The tensor product postulate isn't at all incompatible with classical mechanics. Consider two particles on $\mathbb{R}^3$. In quantum mechanics, the state space for both is $$\mathcal{H} = L^2(\mathbb{R}^3) \otimes L^2(\mathbb{R}^3) \cong L^2(\mathbb{R}^6).$$ When we take the classical limit, the Hilbert space $L^2(\mathbb{R}^n)$ gives the configuration space $\mathbb{R}^n$, and $$\mathbb{R}^3 \oplus \mathbb{R}^3 = \mathbb{R}^6.$$ That is, the rule for combining configuration spaces in classical mechanics is just a limit of the quantum rule, it isn't chosen independently.

knzhou
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  • I wouldn't really rate it as a postulate, to be honest. If the configuration space of two given particles is $C=C_1\times C_2$ then the corresponding state space can be calculated to be $L^2 (C) = L^2(C_1\times C_2) = L^2(C_1)\otimes L^2(C_2)$. There's no need to appeal to any classical limit, either - it's just the right Hilbert space corresponding to the usual classical (combined) configuration space. – Emilio Pisanty Mar 12 '18 at 18:00
  • @EmilioPisanty Sure, I suppose it can go either way! – knzhou Mar 12 '18 at 19:14
  • @EmilioPisanty - By "postulate" I meant that somebody has to decide what is the right way to combine single particles' spaces to get the correct space for the combined system. If we didn't have to account for entangled states (if that phenomenon didn't exist), could we have chosen to represent the space of the combined system with a $\oplus$ rather than a $\otimes$? – Frank Mar 12 '18 at 19:25
  • @Frank That would be inconsistent with the previous postulates. If you couple "the Hilbert space is $L^2$ over the classical configuration space" and "the configuration space of two systems with configuration spaces $C_1$ and $C_2$ is $C=C_1\times C_2$, exactly like it is in classical physics" (itself a completely uncontroversial claim), then you naturally get a tensor-product state space (as a theorem) and you naturally get entanglement. If you insist on a vector sum of state spaces, you get that e.g. the state space of a particle in 2D isn't $L^2$ over $\mathbb R^2$. – Emilio Pisanty Mar 12 '18 at 19:31
  • So, it sounds like entanglement "comes from" the first postulate you give ("the Hilbert space is L2 over the classical configuration space"), since the second one is already in classical physics, which doesn't have entanglement. The first postulate requires $\otimes$ for consistency, and that forces the existence of entanglement? – Frank Mar 12 '18 at 20:14
  • Thank you for your answer and the reference Niels and Chuang! The question has been also on my mind for quite a while. The astonishing thing is that most QM authors listing "QM postulates" don't include it, some others do. – freecharly Mar 12 '18 at 21:45
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I think that the use of the tensor product vector space generated by the tensor product of state vector spaces of subsystems is a distinct postulate added to the other postulates of QM.

PS: For example, I have found it as a distinct postulate in: Valter Moretti, Mathematical Foundations of Quantum Mechanics: An Advanced Short Course, 2016. The author also participates in physics stackexchange.

PPS: I have found some related question here. However, whether the answer asserts that this is a postulate, seems not to be clear.

freecharly
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The basic postulate is this: the description of a system is the set of probability amplitudes for each possible result of measurement of the system.

If a classical system consists of two subcomponents, each of which can admit N possible results of measurements, then we can describe the system with 2N values.

On the other hand, if a quantum system consists of two subcomponents, each of which can admit N possible results of measurements, then we need N x N values, one for each possible combination of single system outcomes.

Imagine a system composed of two distinguishable balls each of which are red, green, or blue. We could, for instance, specify the state vector of one ball as these probability amplitudes:

Red = 0.70

Green = 0.57

Blue = 0.41

so that the probabilities (the squared probability amplitudes) of each color are red, 0.5; green, 0.33; blue, 0.17.

For the two ball system, the classical description would simply consist of three amplitudes for each ball.

But in QM, in order to specify the state of the two ball system we would need to specify a value for each of the 3 x 3 = 9 combinations of outcomes, for example:

red, red = 0.33

green, green = 0.33

blue, blue = 0.33

red, green = 0.33

green, red = 0.33

red, blue = 0.33

blue, red = 0.33

green, blue = 0.33

blue, green = 0.33

Tim Cruz
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